Help Thrid second and first order differential equation

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SUMMARY

The discussion focuses on solving the third-order linear differential equation y''' - y'' - y' + y - x = 0. The equation is identified as having constant coefficients, with the associated homogeneous equation y''' - y'' - y' + y = 0. The characteristic equation r^3 - r^2 - r + 1 = 0 reveals one root at r = 1, allowing for further factorization to find the remaining roots. A particular solution can be approached by assuming a form y(x) = Ax + B, where A and B are constants.

PREREQUISITES
  • Understanding of differential equations, specifically third-order linear equations.
  • Familiarity with characteristic equations and their solutions.
  • Knowledge of homogeneous and particular solutions in differential equations.
  • Basic algebraic manipulation skills for solving polynomial equations.
NEXT STEPS
  • Study the methods for solving third-order linear differential equations with constant coefficients.
  • Learn about the factorization of characteristic polynomials and finding roots.
  • Explore techniques for deriving particular solutions to non-homogeneous differential equations.
  • Practice solving various examples of linear differential equations to reinforce understanding.
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Students preparing for tests in differential equations, educators teaching advanced mathematics, and anyone seeking to enhance their problem-solving skills in linear differential equations.

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Help! Thrid second and first order differential equation!

I have no idea how to accomplish this problem. If anyone knows help please help me solve this example before I take my test!

Solve
y''' - y'' - y' + y - x = 0
 
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Have you never taken a course in differential equations? If not I wonder why you would be trying to solve an equation like this. If you have, this should be relatively easy. It is a "linear equation with constant coefficients". The associated homogeneous equation is y'''- y''- y'+ y= 0 which has characteristic equation [itex]r^3- r^2- r+ 1= 0[/itex]. An obvious solution to that is r= 1 and then you can divide by r- 1 to get a quadratic equation to solve for the other two roots.

To find a specific solution to the entire equation, try y(x)= Ax+ B for some numbers A and B.
 

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